Science:Math Exam Resources/Courses/MATH101/April 2007/Question 05

MATH101 April 2007

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[hide] Question 05
Full-Solution Problem. Justify your answers and show all your work. Simplification of answers is not required. A paper cup has the shape depicted below. All of its horizontal cross sections are circles; the radius of the cup's bottom is 3 cm and the radius of its top is 6 cm. The cup is full of Cona Cola, which has a density of 1000 kg/ $m^{3}$ . The Cona Cola is drunk through a vertical straw that extends 10 cm above the top of the cup and reaches the bottom of the cup. Express as an explicit definite integral the work performed in drinking all the cola. Do not evaluate this integral. For the acceleration due to gravity, use the value $9.8m/s^{2}$ .

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[show] Hint 1
Place a coordinate system on your picture with the top being 0. Then choose a sample point $\displaystyle x_{i}^{*}$ . The best next step is to extend the picture to a cone and relate the sample point to the radius at that point via similar triangles.

[show] Hint 2
The following formulas will probably be useful Volume of a cylinder = pi x radius² x height Mass = volume x density Force = acceleration x mass = gravity x mass = 9.8 x mass Work = force x displacement

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[show] Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Our first step will be to extend the cup to look like a cone and label everything as shown in the diagram. Now, we can compute the value of y quickly via similar triangles (the big triangle and the one with side length 3), ${\frac {20+y}{6}}={\frac {y}{3}}$ and this gives y = 20cm. However, notice in the question that there are several different units everywhere and so for consistency we should use a single unit (metres). As shown, let $x_{i}^{}$ be a sample point between 0 and 0.2 (0 to 20cm), measured from the top of the cup. Then, at this point, the radius, labeled as $r_{i}$ is ${\frac {r_{i}}{0.4-x_{i}^{}}}={\frac {0.06}{0.4}}={\frac {3}{20}}$ and isolating gives $r_{i}={\frac {3(0.4-x_{i}^{})}{20}}={\frac {1.2-3x_{i}^{}}{20}}$ Now, using the volume of a cylinder formula, we have $V_{i}=\pi r_{i}^{2}\Delta x=\pi {\bigg (}{\frac {1.2-3x_{i}^{}}{20}}{\bigg )}^{2}\Delta x$ Using the mass formula with $\delta$ as density, we have $m_{i}=\delta V_{i}=1000\pi {\bigg (}{\frac {1.2-3x_{i}^{}}{20}}{\bigg )}^{2}\Delta x$ Using the force formula, we have $F_{i}=m_{i}g=(9.8)1000\pi {\bigg (}{\frac {1.2-3x_{i}^{}}{20}}{\bigg )}^{2}\Delta x$ and finally, using the work formula, we can compute that the work done at the sample point is $W_{i}=F_{i}d=(9.8)1000\pi {\bigg (}{\frac {1.2-3x_{i}^{}}{20}}{\bigg )}^{2}\Delta x(0.1+x_{i}^{})$ where we added 0.1m (10cm) to the distance d above since we are using a straw that is 10 cm above the cup. Hence, summing all these pieces gives ${\begin{aligned}\lim _{n\to \infty }\sum _{i=1}^{n}W_{i}&=\lim _{n\to \infty }\sum _{i=1}^{n}(9.8)1000\pi {\bigg (}{\frac {1.2-3x_{i}^{}}{20}}{\bigg )}^{2}\Delta x(0.1+x_{i}^{*})\\&=\int _{0}^{20}(9.8)1000\pi {\bigg (}{\frac {1.2-3x}{20}}{\bigg )}^{2}(0.1+x)\,dx\end{aligned}}$ completing the question (we do not need to evaluate this integral).